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Soloha48 [4]
3 years ago
9

why are we all cheating if we payed attention in what they trying to tell us we wouldn't have to cheat but we are not paying att

ention and we have to cheat or we with fail ​
Physics
2 answers:
sweet-ann [11.9K]3 years ago
7 0
I’ve always been failing since middle school. it’s bcs of quarantine that made me unmotivated. rn my grades are F’s D C and A . I should be paying attention but my phone just keeps me distracted lol.
WINSTONCH [101]3 years ago
3 0

honestly what you said is true. but it is very hard to have interest in something like school when we could distract ourselves with something else and avoid paying attention.

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When a closed loop of wire experiences a changing magnetic field a voltage and therefore current is produced. This is the basis
MakcuM [25]

Answer: An electric motor  

Explanation:

I took the quiz for physics and this was the answer

8 0
3 years ago
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Encuentre en km.h la velocidad de un tigre que corre 550 km.h en 69min​
Len [333]

Answer:

v = 478.26 km/h

Explanation:

The question is "find in km.h the speed of a tiger that runs 550 km in 69min"

Distance, d = 550 km

Time, t = 69 min = 1.15 h

We need to find the speed of the tiger. The speed of an object is equal to the total distance covered divided by time taken. So,

v=\dfrac{550\ km}{1.15\ h}\\\\v=478.26\ km/h

So, the speed of the tiger is 478.26 km/h.

8 0
3 years ago
Please help me out! Thanks! (:
ElenaW [278]

X=Ft is a possible equation for X because by using SI units its final term became Kg metre per second

3 0
3 years ago
Two long parallel wires 40 cm apart are carrying currents of 10 A and 20 A in the opposite direction. What is the magnitude of t
Alex_Xolod [135]

Answer:

The magnitude of the magnetic field halfway between the wires is 3.0 x 10⁻⁵ T.

Explanation:

Given;

distance half way between the parallel wires, r = ¹/₂ (40 cm) = 20 cm = 0.2 m

current carried in opposite direction, I₁ and I₂ = 10 A and 20 A respectively

The magnitude of the magnetic field halfway between the wires can be calculated as;

B = \frac{\mu _oI_1}{2 \pi r} + \frac{\mu_oI_2}{2\pi r}

where;

B is magnitude of the magnetic field halfway between the wires

I₁ is current in the first wire

I₂ is current the second wire

μ₀ is permeability of free space

r is distance half way between the wires

B = \frac{\mu_o I_1}{2\pi r} + \frac{\mu_o I_2}{2\pi r} \\\\B = \frac{\mu_o }{2\pi r} (I_1 +I_2)\\\\B = \frac{4\pi *10^{-7} }{2\pi *0.2} (10 +20) = 3.0 *10^{-5}\  T

Therefore, the magnitude of the magnetic field halfway between the wires is 3.0 x 10⁻⁵ T.

5 0
3 years ago
A centrifuge in a biology laboratory rotates at an angular speed of 3,400 rev/min. When switched off, it rotates 48.0 times befo
IgorLugansk [536]

Answer:

- 210 rad/s²

Explanation:

n = frequency of rotation = 3400/60 = 170/3 per sec.

angular velocity  ω ( 0 ) at time 0  = 2π n = 2π x 170/3

angular velocity at time t = ω(t) = 0

now, ω²( t) = w²(o) + 2α Φ ( α = angular acceleration and Φ = angular displacement) = 2π x 48 rad.

0 = ( 2π x 170/3 )² + 2α x 48 x 2π

α = - (2π x 170 x 170 )/ (3 x 3 x 2 x 48 ) = 210 rad / s²

7 0
3 years ago
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